introduction to logic lecture 6 natural...
TRANSCRIPT
![Page 1: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/1.jpg)
INTRODUCTION TO LOGIC
Lecture 6Natural Deduction
Dr. James Studd
There’s nothing you can’t proveif your outlook is only sufficiently limited
Dorothy L. Sayers
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Outline1 Proof2 Rules for connectives3 Rules for quantifiers4 Adequacy
![Page 3: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/3.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusionThe unbracketed sentences at the top are the premissesEach line is an instance of one of 17 rulesThe rules depend purely on the syntax of the sentences. . . not on their semantic properties.
![Page 4: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/4.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusionThe unbracketed sentences at the top are the premissesEach line is an instance of one of 17 rulesThe rules depend purely on the syntax of the sentences. . . not on their semantic properties.
![Page 5: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/5.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusion
The unbracketed sentences at the top are the premissesEach line is an instance of one of 17 rulesThe rules depend purely on the syntax of the sentences. . . not on their semantic properties.
![Page 6: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/6.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusionThe unbracketed sentences at the top are the premisses
Each line is an instance of one of 17 rulesThe rules depend purely on the syntax of the sentences. . . not on their semantic properties.
![Page 7: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/7.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusionThe unbracketed sentences at the top are the premissesEach line is an instance of one of 17 rules
The rules depend purely on the syntax of the sentences. . . not on their semantic properties.
![Page 8: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/8.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusionThe unbracketed sentences at the top are the premissesEach line is an instance of one of 17 rulesThe rules depend purely on the syntax of the sentences
. . . not on their semantic properties.
![Page 9: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/9.jpg)
Proofs
Proofs in Natural DeductionProofs in Natural Deduction are trees of L2-sentences
[Pa]
∀y (Py → Qy)
Pa→ Qa
Qa
∀z (Qz → Rz)
Qa→ Ra
RaPa→ Ra
∀y (Py → Ry)
The root of the tree is the conclusionThe unbracketed sentences at the top are the premissesEach line is an instance of one of 17 rulesThe rules depend purely on the syntax of the sentences. . . not on their semantic properties.
![Page 10: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/10.jpg)
6.1 Propositional logic
Rules for ∧∧IntroThe result of appending φ ∧ ψ to a proof of φ and a proofof ψ is a proof of φ ∧ ψ.
...φ
...ψ
∧Introφ ∧ ψ
∧Elim1 and ∧Elim2(1) The result of appending φ to a proof of φ ∧ ψ is a proof of φ.(2) The result of appending ψ to a proof of φ∧ψ is a proof of ψ.
...φ ∧ ψ
∧Elim1φ
...φ ∧ ψ
∧Elim2ψ
![Page 11: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/11.jpg)
6.1 Propositional logic
Rules for ∧∧IntroThe result of appending φ ∧ ψ to a proof of φ and a proofof ψ is a proof of φ ∧ ψ.
...φ
...ψ
∧Introφ ∧ ψ
∧Elim1 and ∧Elim2(1) The result of appending φ to a proof of φ ∧ ψ is a proof of φ.(2) The result of appending ψ to a proof of φ∧ψ is a proof of ψ.
...φ ∧ ψ
∧Elim1φ
...φ ∧ ψ
∧Elim2ψ
![Page 12: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/12.jpg)
6.1 Propositional logic
Rules for ∧∧IntroThe result of appending φ ∧ ψ to a proof of φ and a proofof ψ is a proof of φ ∧ ψ.
...φ
...ψ
∧Introφ ∧ ψ
∧Elim1 and ∧Elim2(1) The result of appending φ to a proof of φ ∧ ψ is a proof of φ.(2) The result of appending ψ to a proof of φ∧ψ is a proof of ψ.
...φ ∧ ψ
∧Elim1φ
...φ ∧ ψ
∧Elim2ψ
![Page 13: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/13.jpg)
6.1 Propositional logic
Example(P ∧Q) ∧R ` P
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6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧R
P ∧QP
First, assume the premiss.This is covered by the
assumption ruleThe occurrence of a sentence φ withno sentence above it is anassumption. An assumption of φ isa proof of φ.
You may assume any sentence.(But choosing the right assumptionsis important.)
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6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧RP ∧Q
P
Next apply a rule ∧Elim1.
...φ ∧ ψ
∧Elim1φ
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6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧RP ∧QP
Next apply the same rule a secondtime.
...φ ∧ ψ
∧Elim1φ
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6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧RP ∧QP
That’s it! We have a complete proof.
The conclusion is the sentenceat the root.The premiss is the sentence atthe topEach line is a correctapplication of a NaturalDeduction rule.
![Page 18: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/18.jpg)
6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧RP ∧QP
That’s it! We have a complete proof.
The conclusion is the sentenceat the root.
The premiss is the sentence atthe topEach line is a correctapplication of a NaturalDeduction rule.
![Page 19: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/19.jpg)
6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧RP ∧QP
That’s it! We have a complete proof.
The conclusion is the sentenceat the root.The premiss is the sentence atthe top
Each line is a correctapplication of a NaturalDeduction rule.
![Page 20: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/20.jpg)
6.1 Propositional logic
Example(P ∧Q) ∧R ` P
(P ∧Q) ∧RP ∧QP
That’s it! We have a complete proof.
The conclusion is the sentenceat the root.The premiss is the sentence atthe topEach line is a correctapplication of a NaturalDeduction rule.
![Page 21: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/21.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
...φ
...ψ
∧Introφ ∧ ψ
...φ ∧ ψ
∧Elim1φ
...φ ∧ ψ
∧Elim2ψ
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6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
...φ
...ψ
∧Introφ ∧ ψ
...φ ∧ ψ
∧Elim1φ
...φ ∧ ψ
∧Elim2ψ
![Page 23: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/23.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ Pa
Pa RaPa ∧Ra
First assume the first premiss.
![Page 24: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/24.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa
RaPa ∧Ra
Next apply a rule for ∧.
...φ ∧ ψ
∧Elim2ψ
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6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa Ra
Pa ∧Ra
Now assume the second premiss
![Page 26: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/26.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa RaPa ∧Ra
Now apply the introduction rule for∧.
...φ
...ψ
∧Introφ ∧ ψ
![Page 27: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/27.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa RaPa ∧Ra
The proof is complete.
The conclusion is the sentenceat the root.The premisses are thesentences at the top.Each line is a correctapplication of a NaturalDeduction rule.
![Page 28: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/28.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa RaPa ∧Ra
The proof is complete.The conclusion is the sentenceat the root.
The premisses are thesentences at the top.Each line is a correctapplication of a NaturalDeduction rule.
![Page 29: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/29.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa RaPa ∧Ra
The proof is complete.The conclusion is the sentenceat the root.The premisses are thesentences at the top.
Each line is a correctapplication of a NaturalDeduction rule.
![Page 30: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/30.jpg)
6.1 Propositional logic
ExampleQb ∧ Pa,Ra ` Pa ∧Ra
Qb ∧ PaPa RaPa ∧Ra
The proof is complete.The conclusion is the sentenceat the root.The premisses are thesentences at the top.Each line is a correctapplication of a NaturalDeduction rule.
![Page 31: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/31.jpg)
6.1 Propositional logic
Rules for →→ElimThe result of appending ψ to a proof of φ and a proof of φ→ ψ isa proof of ψ.
...φ
...φ→ ψ
→Elimψ
This rule is often called ‘Modus Ponens’.
![Page 32: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/32.jpg)
6.1 Propositional logic
Rules for →→ElimThe result of appending ψ to a proof of φ and a proof of φ→ ψ isa proof of ψ.
...φ
...φ→ ψ
→Elimψ
This rule is often called ‘Modus Ponens’.
![Page 33: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/33.jpg)
6.1 Propositional logic
Rules for →→ElimThe result of appending ψ to a proof of φ and a proof of φ→ ψ isa proof of ψ.
...φ
...φ→ ψ
→Elimψ
This rule is often called ‘Modus Ponens’.
![Page 34: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/34.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
...φ
...φ→ ψ
→Elimψ
![Page 35: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/35.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
...φ
...φ→ ψ
→Elimψ
![Page 36: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/36.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
∃y Py ∃y Py → Qa
Qa
Assume both premisses.
![Page 37: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/37.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
∃y Py ∃y Py → QaQa
Apply the eliminationrule.
...φ
...φ→ ψ
→Elimψ
![Page 38: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/38.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
∃y Py ∃y Py → QaQa
Finished!
The conclusion is thesentence at the root.The only assumptionsare premisses.Each line is a correctapplication of aNatural Deductionrule.
![Page 39: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/39.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
∃y Py ∃y Py → QaQa
Finished!The conclusion is thesentence at the root.
The only assumptionsare premisses.Each line is a correctapplication of aNatural Deductionrule.
![Page 40: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/40.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
∃y Py ∃y Py → QaQa
Finished!The conclusion is thesentence at the root.The only assumptionsare premisses.
Each line is a correctapplication of aNatural Deductionrule.
![Page 41: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/41.jpg)
6.1 Propositional logic
Example∃y Py → Qa,∃y Py ` Qa
∃y Py ∃y Py → QaQa
Finished!The conclusion is thesentence at the root.The only assumptionsare premisses.Each line is a correctapplication of aNatural Deductionrule.
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6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 43: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/43.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 44: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/44.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 45: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/45.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof.
Assume Quintus took it.Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 46: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/46.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.
Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 47: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/47.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.Then (by 2) it’s poison and he took it.
Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 48: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/48.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 49: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/49.jpg)
6.1 Propositional logic
→IntroThe result of appending φ→ ψ to a proof of ψ and discharging allassumptions of φ in the proof of ψ is a proof of φ→ ψ.
[φ]
...ψ
→Introφ→ ψ
Conditional proof in informal reasoning.(1) If it’s poison and Quintus took it, then he needs to be readmitted.(2) It’s poisonSo (C) if Quintus took it, he need to be readmitted.
Informal proof. Assume Quintus took it.Then (by 2) it’s poison and he took it.Then (by 1 and MP) he needs to be readmitted.
So (by conditional proof) if Quintus took it, he needs to be readmitted.
![Page 50: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/50.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R 25
[φ]
...ψ
→Introφ→ ψ
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6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R 25
[φ]
...ψ
→Introφ→ ψ
![Page 52: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/52.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q (P ∧Q) → R
RQ→ R
Assume the first premiss
![Page 53: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/53.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q (P ∧Q) → R
RQ→ R
Next assume Q.This is the standardway to prove aconditional conclusion.We assume theantecedent and provethe consequent.
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6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q
(P ∧Q) → R
RQ→ R
Apply ∧Intro.
...φ
...ψ
∧Introφ ∧ ψ
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6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q (P ∧Q) → R
RQ→ R
Assume the secondpremiss.
![Page 56: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/56.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q (P ∧Q) → RR
Q→ R
Apply →Elim.
...φ
...φ→ ψ
→Elimψ
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6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q (P ∧Q) → RR
Q→ R
Assuming the antecedentQ we’ve reached theconsequent R.
So we may apply →Intro
[φ]
...ψ
→Introφ→ ψ
We discharge theassumption of Q.
![Page 58: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/58.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P
[
Q
]
P ∧Q (P ∧Q) → RR
Q→ R
Assuming the antecedentQ we’ve reached theconsequent R.
So we may apply →Intro
[φ]
...ψ
→Introφ→ ψ
We discharge theassumption of Q.
![Page 59: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/59.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P [Q]P ∧Q (P ∧Q) → R
RQ→ R
Assuming the antecedentQ we’ve reached theconsequent R.
So we may apply →Intro
[φ]
...ψ
→Introφ→ ψ
We discharge theassumption of Q.
![Page 60: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/60.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P [Q]P ∧Q (P ∧Q) → R
RQ→ R
The proof is complete
The conclusion is atthe root.The only undischargedassumptions arepremisses.Dischargedassumptions don’tneed to be amongstthe premisses.
![Page 61: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/61.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P [Q]P ∧Q (P ∧Q) → R
RQ→ R
The proof is completeThe conclusion is atthe root.
The only undischargedassumptions arepremisses.Dischargedassumptions don’tneed to be amongstthe premisses.
![Page 62: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/62.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P [Q]P ∧Q (P ∧Q) → R
RQ→ R
The proof is completeThe conclusion is atthe root.The only undischargedassumptions arepremisses.
Dischargedassumptions don’tneed to be amongstthe premisses.
![Page 63: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/63.jpg)
6.1 Propositional logic
ExampleP, (P ∧Q) → R ` Q→ R
P [Q]P ∧Q (P ∧Q) → R
RQ→ R
The proof is completeThe conclusion is atthe root.The only undischargedassumptions arepremisses.Dischargedassumptions don’tneed to be amongstthe premisses.
![Page 64: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/64.jpg)
6.1 Propositional logic
We can now define Γ ` φ.
Let Γ be a set of sentences and φ a sentence.
Definition (Provable)The sentence φ is provable from Γ if and only if:
there is a proof of φ with only sentences in Γ asnon-discharged assumptions.
NotationΓ ` φ is short for φ is provable from Γ
` φ is short for ∅ ` φψ1, . . . , ψn ` φ is short for {ψ1, . . . , ψn} ` φ.
![Page 65: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/65.jpg)
6.1 Propositional logic
We can now define Γ ` φ.Let Γ be a set of sentences and φ a sentence.
Definition (Provable)The sentence φ is provable from Γ if and only if:
there is a proof of φ with only sentences in Γ asnon-discharged assumptions.
NotationΓ ` φ is short for φ is provable from Γ
` φ is short for ∅ ` φψ1, . . . , ψn ` φ is short for {ψ1, . . . , ψn} ` φ.
![Page 66: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/66.jpg)
6.1 Propositional logic
We can now define Γ ` φ.Let Γ be a set of sentences and φ a sentence.
Definition (Provable)The sentence φ is provable from Γ if and only if:
there is a proof of φ with only sentences in Γ asnon-discharged assumptions.
NotationΓ ` φ is short for φ is provable from Γ
` φ is short for ∅ ` φψ1, . . . , ψn ` φ is short for {ψ1, . . . , ψn} ` φ.
![Page 67: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/67.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 68: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/68.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 69: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/69.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.
Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 70: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/70.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 71: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/71.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 72: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/72.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 73: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/73.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 74: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/74.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 75: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/75.jpg)
6.1 Propositional logic
Return to the rule of assumption.
assumption ruleThe occurrence of a sentence φ with no sentence above it isan assumption. An assumption of φ is a proof of φ.
This may seem odd.Suppose I assume, the following:
∃x∃y(Rxy ∨ P )
By the rule, this counts as a proof of ∃x∃y(Rxy ∨ P )
But it is not an outright proof of ∃x∃y(Rxy ∨ P )
This proof does not show ` ∃x∃y(Rxy ∨ P )
Instead it shows ∃x∃y(Rxy ∨ P ) ` ∃x∃y(Rxy ∨ P )
![Page 76: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/76.jpg)
6.1 Propositional logic
Rules for ∨The introduction rules are straightforward.
...φ
∨Intro1φ ∨ ψ
...ψ
∨Intro2φ ∨ ψ
![Page 77: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/77.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 78: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/78.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 79: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/79.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 80: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/80.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit.
So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 81: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/81.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 82: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/82.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play.
So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 83: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/83.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 84: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/84.jpg)
6.1 Propositional logic
The elimination rule is a little more complex.
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
Proof by cases in informal reasoning(1) Either you don’t play and you quit or you do something quietand don’t play.So, (C) you don’t play.
Informal proof. Suppose (1)
Case (i): You don’t play and you quit. So: you don’t play
Case (ii): You do something quiet and don’t play. So: you don’tplay.
Either way then, (C) follows: you don’t play.
![Page 85: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/85.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
![Page 86: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/86.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
![Page 87: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/87.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P¬P
Assume the premiss
![Page 88: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/88.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P¬P
Case 1: Assume ¬P ∧Q
![Page 89: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/89.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P¬P
Apply ∧Elim1
...φ ∧ ψ
∧Elim1φ
That completes case 1.
![Page 90: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/90.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P¬P
Case 2: Assume ∃xQx ∧ ¬P
![Page 91: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/91.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P
¬P
Apply ∧Elim1 once more
...φ ∧ ψ
∧Elim1φ
That completes case 2.
![Page 92: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/92.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P
¬P
We’ve reached ¬P from each disjunct.We can now apply ∨Elim
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
![Page 93: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/93.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[
¬P ∧Q
]
¬P
[
∃xQx ∧ ¬P
]
¬P¬P
We’ve reached ¬P from each disjunct.We can now apply ∨Elim
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
![Page 94: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/94.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[¬P ∧Q]
¬P
[
∃xQx ∧ ¬P
]
¬P¬P
We’ve reached ¬P from each disjunct.We can now apply ∨Elim
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
![Page 95: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/95.jpg)
6.1 Propositional logic
Example(¬P ∧Q) ∨ (∃xQx ∧ ¬P ) ` ¬P
(¬P ∧Q) ∨ (∃xQx ∧ ¬P )
[¬P ∧Q]
¬P[∃xQx ∧ ¬P ]
¬P¬P
We’ve reached ¬P from each disjunct.We can now apply ∨Elim
...φ ∨ ψ
[φ]
...χ
[ψ]
...χ
∨Elimχ
![Page 96: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/96.jpg)
6.1 Propositional logic
The rules for ¬Here are the rules for ¬.
[φ]
...ψ
[φ]
...¬ψ
¬Intro¬φ
[¬φ]
...ψ
[¬φ]
...¬ψ
¬Elimφ
The proof technique is known as reductio ad absurdum.
![Page 97: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/97.jpg)
6.1 Propositional logic
The rules for ¬Here are the rules for ¬.
[φ]
...ψ
[φ]
...¬ψ
¬Intro¬φ
[¬φ]
...ψ
[¬φ]
...¬ψ
¬Elimφ
The proof technique is known as reductio ad absurdum.
![Page 98: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/98.jpg)
6.1 Propositional logic
Example¬(P → Q) ` ¬Q
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6.1 Propositional logic
Example¬(P → Q) ` ¬Q
[
Q
]
P → Q ¬(P → Q)
¬Q
[¬φ]
...ψ
[¬φ]
...¬ψ
¬Elimφ
[φ]
...ψ
[φ]
...¬ψ
¬Intro¬φ
![Page 100: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/100.jpg)
6.1 Propositional logic
Example¬(P → Q) ` ¬Q
[
Q
]
P → Q ¬(P → Q)
¬Q
We want to prove ¬Q be reductio.So start by assuming Q, and we’ll go for a contradiction.
![Page 101: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/101.jpg)
6.1 Propositional logic
Example¬(P → Q) ` ¬Q
[
Q
]
P → Q
¬(P → Q)
¬Q
We can always safely assume the premiss.
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6.1 Propositional logic
Example¬(P → Q) ` ¬Q
[
Q
]
P → Q ¬(P → Q)
¬Q
Next apply →Intro to get a contradiction
[φ]
...ψ
→Introφ→ ψ
(Note this rule can be applied even when we haven’t assumed theantecedent)
![Page 103: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/103.jpg)
6.1 Propositional logic
Example¬(P → Q) ` ¬Q
[
Q
]
P → Q ¬(P → Q)
¬QNow we apply ¬Intro
[φ]
...ψ
[φ]
...¬ψ
¬Intro¬φ
And we’re done.
![Page 104: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/104.jpg)
6.1 Propositional logic
Example¬(P → Q) ` ¬Q
[Q]
P → Q ¬(P → Q)
¬QNow we apply ¬Intro
[φ]
...ψ
[φ]
...¬ψ
¬Intro¬φ
And we’re done.
![Page 105: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/105.jpg)
6.1 Propositional logic
Rules for ↔These are reminiscent of the rules for →
[φ]
...ψ
[ψ]
...φ
↔Introφ↔ ψ
...φ↔ ψ
...φ
↔Elim1ψ
...φ↔ ψ
...ψ
↔Elim2φ
![Page 106: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/106.jpg)
6.2 Predicate logic
Rules for ∀...
∀v φ∀Elim
φ[t/v]
In this rule:φ is a formula in which only the variable v occurs freelyt is a constantφ[t/v] is the sentence obtained by replacing all freeoccurrences of v in φ by t.
![Page 107: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/107.jpg)
6.2 Predicate logic
Rules for ∀...
∀v φ∀Elim
φ[t/v]
In this rule:φ is a formula in which only the variable v occurs freelyt is a constantφ[t/v] is the sentence obtained by replacing all freeoccurrences of v in φ by t.
![Page 108: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/108.jpg)
6.2 Predicate logic
Substitutionφ[t/v] is the sentence obtained by replacing all freeoccurrences of v in φ by t.
Recall that a free occurrence of v is one not bound by ∃v
Compute the followingPx[a/x] =
Pa
∀xPx[a/x] =
∀xPx
∀y(∃xPx ∨Qx→ Py)[a/x] =
∀y(∃xPx ∨Qa→ Py)
40
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6.2 Predicate logic
Substitutionφ[t/v] is the sentence obtained by replacing all freeoccurrences of v in φ by t.
Recall that a free occurrence of v is one not bound by ∃v
Compute the followingPx[a/x] = Pa
∀xPx[a/x] =
∀xPx
∀y(∃xPx ∨Qx→ Py)[a/x] =
∀y(∃xPx ∨Qa→ Py)
40
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6.2 Predicate logic
Substitutionφ[t/v] is the sentence obtained by replacing all freeoccurrences of v in φ by t.
Recall that a free occurrence of v is one not bound by ∃v
Compute the followingPx[a/x] = Pa
∀xPx[a/x] = ∀xPx∀y(∃xPx ∨Qx→ Py)[a/x] =
∀y(∃xPx ∨Qa→ Py)
40
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6.2 Predicate logic
Substitutionφ[t/v] is the sentence obtained by replacing all freeoccurrences of v in φ by t.
Recall that a free occurrence of v is one not bound by ∃v
Compute the followingPx[a/x] = Pa
∀xPx[a/x] = ∀xPx∀y(∃xPx ∨Qx→ Py)[a/x] = ∀y(∃xPx ∨Qa→ Py) 40
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
...∀v φ
∀Elimφ[t/v]
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
...∀v φ
∀Elimφ[t/v]
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
Pa
∀x (Px→ Qx)
Pa→ Qa
Qa
Assume the first premiss.
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
Pa
∀x (Px→ Qx)
Pa→ Qa
Qa
Apply ∀Elim
...∀v φ
∀Elimφ[t/v]
To apply the rule: delete ∀x and byreplace all occurrences of x in the formulaby the constant a.
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
Pa
∀x (Px→ Qx)
Pa→ Qa
Qa
Assume the other premiss
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
Pa
∀x (Px→ Qx)
Pa→ QaQa
Apply modus ponens.
...φ
...φ→ ψ
→Elimψ
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6.2 Predicate logic
Example∀x (Px→ Qx), Pa ` Qa
Pa
∀x (Px→ Qx)
Pa→ QaQa
And we’re done
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6.2 Predicate logic
Here’s the introduction rule for ∀
...φ[t/v]
∀Intro∀v φ
side conditions:
(i) the constant t does not occurin φ and
(ii) t does not occur in anyundischarged assumption in theproof of φ[t/v].
Informal reasoning with arbitrary names(C) Every person is either a quorn-lover or a person
Informal proof. Let an arbitrary thing be given.Call it ‘Jane Doe’.
Clearly, if Jane Doe is a person, then Jane Doe is either aquorn-lover or a person.
So: every person is either a quorn-lover or a person.
![Page 120: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/120.jpg)
6.2 Predicate logic
Here’s the introduction rule for ∀
...φ[t/v]
∀Intro∀v φ
side conditions:
(i) the constant t does not occurin φ and
(ii) t does not occur in anyundischarged assumption in theproof of φ[t/v].
Informal reasoning with arbitrary names(C) Every person is either a quorn-lover or a person
Informal proof. Let an arbitrary thing be given.Call it ‘Jane Doe’.
Clearly, if Jane Doe is a person, then Jane Doe is either aquorn-lover or a person.
So: every person is either a quorn-lover or a person.
![Page 121: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/121.jpg)
6.2 Predicate logic
Here’s the introduction rule for ∀
...φ[t/v]
∀Intro∀v φ
side conditions:
(i) the constant t does not occurin φ and
(ii) t does not occur in anyundischarged assumption in theproof of φ[t/v].
Informal reasoning with arbitrary names(C) Every person is either a quorn-lover or a person
Informal proof.
Let an arbitrary thing be given.Call it ‘Jane Doe’.
Clearly, if Jane Doe is a person, then Jane Doe is either aquorn-lover or a person.
So: every person is either a quorn-lover or a person.
![Page 122: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/122.jpg)
6.2 Predicate logic
Here’s the introduction rule for ∀
...φ[t/v]
∀Intro∀v φ
side conditions:
(i) the constant t does not occurin φ and
(ii) t does not occur in anyundischarged assumption in theproof of φ[t/v].
Informal reasoning with arbitrary names(C) Every person is either a quorn-lover or a person
Informal proof. Let an arbitrary thing be given.Call it ‘Jane Doe’.
Clearly, if Jane Doe is a person, then Jane Doe is either aquorn-lover or a person.
So: every person is either a quorn-lover or a person.
![Page 123: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/123.jpg)
6.2 Predicate logic
Here’s the introduction rule for ∀
...φ[t/v]
∀Intro∀v φ
side conditions:
(i) the constant t does not occurin φ and
(ii) t does not occur in anyundischarged assumption in theproof of φ[t/v].
Informal reasoning with arbitrary names(C) Every person is either a quorn-lover or a person
Informal proof. Let an arbitrary thing be given.Call it ‘Jane Doe’.
Clearly, if Jane Doe is a person, then Jane Doe is either aquorn-lover or a person.
So: every person is either a quorn-lover or a person.
![Page 124: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/124.jpg)
6.2 Predicate logic
Here’s the introduction rule for ∀
...φ[t/v]
∀Intro∀v φ
side conditions:
(i) the constant t does not occurin φ and
(ii) t does not occur in anyundischarged assumption in theproof of φ[t/v].
Informal reasoning with arbitrary names(C) Every person is either a quorn-lover or a person
Informal proof. Let an arbitrary thing be given.Call it ‘Jane Doe’.
Clearly, if Jane Doe is a person, then Jane Doe is either aquorn-lover or a person.
So: every person is either a quorn-lover or a person.
![Page 125: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/125.jpg)
6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[
Pa
]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
...φ[t/v]
∀Intro∀v φ
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[
Pa
]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
I assume Pa.(We’ll try to prove Pa→ Qa ∨ Pawithout making assumptions about a)
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[
Pa
]
Qa ∨ Pa
Pa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Apply ∨Intro2.
...ψ
∨Intro2φ ∨ ψ
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[
Pa
]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Apply →Intro
[φ]
...ψ
→Introφ→ ψ
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[Pa]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Apply →Intro
[φ]
...ψ
→Introφ→ ψ
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[Pa]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Finally we want to apply the rule forintroducing ∀.
...φ[t/v]
∀Intro∀v φ
But we also need to check the sideconditions are met.for t = a; φ = (Pz → (Qz ∨ Pz))
(i) t does not occur in φ
i.e. a does not occur in(Pz → (Qz ∨ Pz))
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6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[Pa]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Finally we want to apply the rule forintroducing ∀.
...φ[t/v]
∀Intro∀v φ
But we also need to check the sideconditions are met.for t = a; φ = (Pz → (Qz ∨ Pz))
(i) t does not occur in φ
i.e. a does not occur in(Pz → (Qz ∨ Pz))
![Page 133: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/133.jpg)
6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[Pa]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Finally we want to apply the rule forintroducing ∀.
...φ[t/v]
∀Intro∀v φ
But we also need to check the sideconditions are met.for t = a; φ = (Pz → (Qz ∨ Pz))
(i) t does not occur in φ
i.e. a does not occur in(Pz → (Qz ∨ Pz))
(ii) t does not occur in any undischargedassumption in the proof of φ[t/v].
i.e. a does not occur in undischargedassumptions.
![Page 134: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/134.jpg)
6.2 Predicate logic
Example` ∀z (Pz → Qz ∨ Pz)
[Pa]
Qa ∨ PaPa→ (Qa ∨ Pa)
∀z (Pz → (Qz ∨ Pz))
Finally we want to apply the rule forintroducing ∀.
...φ[t/v]
∀Intro∀v φ
But we also need to check the sideconditions are met.for t = a; φ = (Pz → (Qz ∨ Pz))
(ii) t does not occur in any undischargedassumption in the proof of φ[t/v].
i.e. a does not occur in undischargedassumptions.
![Page 135: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/135.jpg)
6.2 Predicate logic
Rules for ∃The introduction rule is straightforward.
φ[t/v]∃Intro∃v φ
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6.2 Predicate logic
ExampleRcc ` ∃y Rcy
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6.2 Predicate logic
ExampleRcc ` ∃y Rcy
Rcc
∃y Rcy
Assume the premiss.
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6.2 Predicate logic
ExampleRcc ` ∃y Rcy
Rcc
∃y Rcy
Apply ∃Intro
φ[t/v]∃Intro∃v φ
All we need to do is to choose theright φ and v
Let φ = Rcy, v = yφ[c/y] = Rcc∃vφ = ∃yRcy
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6.2 Predicate logic
ExampleRcc ` ∃y Rcy
Rcc∃y Rcy
Apply ∃Intro
φ[t/v]∃Intro∃v φ
All we need to do is to choose theright φ and vLet φ = Rcy, v = yφ[c/y] = Rcc∃vφ = ∃yRcy
![Page 140: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/140.jpg)
6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof. Let Smith be an Albanian penny.
By (2), Smith is a quindarka.So, something is a quindarka.
So (C), follows from (1) and (2).
![Page 141: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/141.jpg)
6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof. Let Smith be an Albanian penny.
By (2), Smith is a quindarka.So, something is a quindarka.
So (C), follows from (1) and (2).
![Page 142: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/142.jpg)
6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof.
Let Smith be an Albanian penny.
By (2), Smith is a quindarka.So, something is a quindarka.
So (C), follows from (1) and (2).
![Page 143: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/143.jpg)
6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof. Let Smith be an Albanian penny.
By (2), Smith is a quindarka.So, something is a quindarka.
So (C), follows from (1) and (2).
![Page 144: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/144.jpg)
6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof. Let Smith be an Albanian penny.
By (2), Smith is a quindarka.
So, something is a quindarka.
So (C), follows from (1) and (2).
![Page 145: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/145.jpg)
6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof. Let Smith be an Albanian penny.
By (2), Smith is a quindarka.So, something is a quindarka.
So (C), follows from (1) and (2).
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6.2 Predicate logic
The elimination rule is as follows.
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
Dummy names again(1) Something is an Albanian penny. (2) Every Albanian penny isa quindarka. So, (C) something is a quindarka.
Informal Proof. Let Smith be an Albanian penny.
By (2), Smith is a quindarka.So, something is a quindarka.
So (C), follows from (1) and (2).
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
The standard way to reason from∃xPx is to assume Pt (for t anew constant)
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
Assume the second premiss.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
Apply ∀Elim.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
Apply →Elim.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx
∃xQx
Apply ∃Intro.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx
∃xQx
Now we’ve reached a conclusionassuming Pc (and making noother assumptions about c) wecan apply ∃Elim.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in theproof of ψ.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) c does not occur in ∃xPx
(ii) t does not occur in ψ
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in theproof of ψ.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) c does not occur in ∃xPx
(ii) t does not occur in ψ
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in theproof of ψ.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) c does not occur in ∃xPx
(ii) c does not occur in ∃xQx
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in theproof of ψ.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) c does not occur in ∃xPx
(ii) c does not occur in ∃xQx
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in theproof of ψ.
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6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[
Pc
]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) c does not occur in ∃xPx
(ii) c does not occur in ∃xQx
(iii) c does not occur in anyundischarged assumptionother than Pc in theproof of ∃xQx.
![Page 160: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/160.jpg)
6.2 Predicate logic
Example∃xPx,∀x (Px→ Qx) ` ∃xQx
∃xPx
[Pc]
∀x (Px→ Qx)
Pc→ Qc
Qc
∃xQx∃xQx
...∃v φ
[φ[t/v]]
...ψ
∃Elimψ
(i) c does not occur in ∃xPx
(ii) c does not occur in ∃xQx
(iii) c does not occur in anyundischarged assumptionother than Pc in theproof of ∃xQx.
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Adequacy
Let Γ be a set of L2-sentences and φ a L2-sentence.
Two notions of consequenceΓ ` φ iff there is a proof of φ with only sentences in Γ asnon-discharged assumptions.
Γ � φ iff there is no L2-structure in which all sentences in Γare true and φ is false.
Theorem(a) Soundness: Γ ` φ only if Γ � φ
(b) Completeness: Γ � φ only if Γ ` φ X
Proof. Elements of Deductive Logic.
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Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax
[
Raa
]
∃xRxx∃xRxx
But clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 163: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/163.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
X t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax[
Raa
]∃xRxx
∃xRxxBut clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 164: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/164.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
X t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax[
Raa
]
∃xRxx
∃xRxxBut clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 165: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/165.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
X t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax
[
Raa
]
∃xRxx
∃xRxxBut clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 166: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/166.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
X t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax
[
Raa
]
∃xRxx∃xRxx
But clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 167: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/167.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
X t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax[Raa]∃xRxx
∃xRxx
But clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 168: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/168.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
X t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xRax[Raa]∃xRxx
∃xRxxBut clearly, ∃xRax 6� ∃xRxx. Without (i), ND is not sound.
![Page 169: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/169.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
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Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
X t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
![Page 171: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/171.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
X t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx [
Pa
]Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
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Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
X t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
![Page 173: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/173.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
X t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
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Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
X t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx [Pa]Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
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Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
X t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx [Pa]Pa
But clearly, ∃xPx 6� Pa. Without (ii), ND is not sound.
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Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
(iii) t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
Qa
Pa ∧Qa∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 177: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/177.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
Qa
Pa ∧Qa∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 178: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/178.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
] Qa
Pa ∧Qa∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 179: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/179.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
Qa
Pa ∧Qa∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 180: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/180.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
QaPa ∧Qa
∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 181: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/181.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
QaPa ∧Qa
∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 182: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/182.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
QaPa ∧Qa
∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 183: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/183.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[
Pa
]
QaPa ∧Qa
∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 184: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/184.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[Pa] QaPa ∧Qa
∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.
![Page 185: INTRODUCTION TO LOGIC Lecture 6 Natural Deductionlogicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf · INTRODUCTION TO LOGIC Lecture 6 Natural Deduction Dr.JamesStudd There’snothingyoucan’tprove](https://reader033.vdocuments.net/reader033/viewer/2022053000/5f04746a7e708231d40e0ce7/html5/thumbnails/185.jpg)
Adequacy
Why we need the side conditions on ∃Elim
...∃v φ
[φ[t/v]]...ψ
∃Elimψ
Side conditions:
(i) t does not occur in ∃v φ
(ii) t does not occur in ψ,
X t does not occur in anyundischarged assumptionother than φ[t/v] in the proofof ψ.
∃xPx
[Pa] QaPa ∧Qa
∃x(Px ∧Qx)
∃x(Px ∧Qx)
But ∃xPx,Qa 6� ∃x(Px ∧Qx). Without (iii), ND is not sound.